OBLICZ MIEJSCE ZEROWE FUNKCJI
f(x) = -3x^2 -2x +1 = 0
miejsca zerowe f kwadratowej a wiec liczymy
a= -3 b= -2 c=1
Δ=b² -4ac
Δ=(-2)²-4(-3)·1= 4+12=16
√Δ=4
sa dwa m zerowe
x₁= -b+√Δ
2a
x₂= -b-√Δ
x₁= 2+4
2(-3)
x₁= -1
x₂= 2- 4
x₂= -2/(-6)=⅓
f(x) = -3x²-2x+1
f(x) = 0
-3x²-2x+1 = 0 I*(-1)
3x²+2x-1 = 0
a = 3, b = 2, c = -1
Δ = b²-4ac = 4+12 = 16
√Δ = √16 = 4
x₁ = (-b-√Δ)/2a = (-2-4)/6 = -6/6 = -1
x₂ = (-b+√Δ)/2a = (-2+6)/6 = 2/6 = 1/3
MZ: -1; 1/3
========
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
miejsca zerowe f kwadratowej a wiec liczymy
a= -3 b= -2 c=1
Δ=b² -4ac
Δ=(-2)²-4(-3)·1= 4+12=16
√Δ=4
sa dwa m zerowe
x₁= -b+√Δ
2a
x₂= -b-√Δ
2a
x₁= 2+4
2(-3)
x₁= -1
x₂= 2- 4
2(-3)
x₂= -2/(-6)=⅓
f(x) = -3x²-2x+1
f(x) = 0
-3x²-2x+1 = 0 I*(-1)
3x²+2x-1 = 0
a = 3, b = 2, c = -1
Δ = b²-4ac = 4+12 = 16
√Δ = √16 = 4
x₁ = (-b-√Δ)/2a = (-2-4)/6 = -6/6 = -1
x₂ = (-b+√Δ)/2a = (-2+6)/6 = 2/6 = 1/3
MZ: -1; 1/3
========